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Trigonometry
Analytical Geometry examples
Example 1 of Ellipse Equation
The standard form of an ellipse is:
[(x - h)
2
)/a
2
+ [(y - k)
2
)/b
2
= 1
the center of the ellipse is (h, k)
and its graph is:
Example 2 of Ellipse Equation
Question 1.
Find the center of the ellipse given by the equation 9x
2
- 36x + 4y
2
+ 24y + 36 = 0 and draw the ellipse.
Solution:
change the given ellipse equation into the stand ellipse equation
9x
2
- 36x + 4y
2
+ 24y + 36 = 0
9(x
2
- 4x) + 4(y
2
+ 6y) + 36 = 0
9[x
2
- 4x + (4/2)
2
- (4/2)
2
] + 4[y
2
+ 6y + (6/2)
2
- (6/2)
2
] + 36 = 0
9[x
2
- 4x + 4 - 4] + 4[y
2
+ 6y + 9 - 9] + 36 = 0
9[x
2
- 4x + 4] - 9 × 4 + 4[y
2
+ 6y + 9] - 4 × 9 + 36 = 0
9[x
2
- 4x + 2
2
] - 36 + 4[y
2
+ 6y + 3
2
]
- 36
+
36
= 0
9[x
2
- 4x + 2
2
] + 4[y
2
+ 6y + 3
2
] = 36
both side of the equation divide by 36 (each item divide by 36)
9[x
2
- 4x + 2
2
]/36 + 4[y
2
+ 6y + 3
2
]/36 = 36/36
[x
2
- 4x + 2
2
]/4 + [y
2
+ 6y + 3
2
]/9 = 1
[(x - 2)
2
]/2
2
+ [(y + 3)
2
]/3
2
= 1
the standard ellipse equation is: [(x - 2)
2
]/2
2
+ [(y + 3)
2
]/3
2
= 1
therefore, the center of the given ellipse is: (2 , - 3)
The graphic of the ellipse is
Example 3 of Circle Equation
Question 2
Find the center of the circle x
2
- 4x + y
2
+ 2y - 4 = 0, draw the graph and find its area.
Solution
x
2
- 4x + y
2
+ 2y - 4 = 0
x
2
- 4x + (4/2)
2
- (4/2)
2
+ y
2
+ 2y + (2/2)
2
- (2/2)
2
- 4 = 0
x
2
- 4x + (2)
2
- (2)
2
+ y
2
+ 2y + (1)
2
- (1)
2
- 4 = 0
x
2
- 4x + 4 - 4 + y
2
+ 2y + 1 - 1 - 4 = 0
x
2
- 4x + 4 + y
2
+ 2y + 1 - 4 - 1 - 4 = 0
(x
2
- 4x + 4) + (y
2
+ 2y + 1) - 4 - 1 - 4 = 0
(x
2
- 4x + 4) + (y
2
+ 2y + 1) = 4 + 1 + 4
(x
2
- 4x + 4) + (y
2
+ 2y + 1) = 9
(x - 2)
2
+ (y + 1)
2
= 3
2
therefore, the center of the circle is: (2, -1) and the radius of the circle is: 3
The center of the circle is (2, -1), and the radius of the circle is r = 3
The area of the circle is pi*r
2
= 3.14 * 9 = 28.26 square units.
remark:
circle equation: (x - x
o
)
2
+ (y - y
o
)
2
= r
2
center of the circle is: (x
o
, y
o
) and radius is r
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